Stoichiometry-the calculation of quantitative (measurable) relationships of the reactants and products in a balanced chemical reaction.
Cookie Example:
You have:
2 cups flour,
3 cups sugar
1/2 tsp vanilla
2 sticks butter
3 cups chocolate chips
These ingredients are for a total of 4 dozen (48 cookies)
What if you want 12 dozen cookies?
You have:
6 cups four
9 cups sugar
1 1/2 tsp vanilla
6 sticks butter
9 cups chocolate chips
These ingredients are for a total of 12 dozen (114 cookies)
This example shows that because each total differs, the amount of ingredients needed also differs. This is similar to stoichiometry in the way that to get the total amount needed you need to change the number of each element.
Steps To Stoichiometry:
Step up an equation.
Balance the equation.
Apply the multipilication factor.
Determine the number of grams using molar mass
You can also download our stoichiometry calculator here:
To download, select save, unzip the file, and double click on set up.
This program will not solve your stoichiometry calculations unless you know what you are doing.
It does not use any form of generic A → B equation, instead A = what you have and B = what you want.
Example:
Use this equation: 2H2O → 2H2 + O2
You have 400 grams of water. How many grams of Oxygen is left over?
To solve this equation, you must use stoichiometry. The grams to grams portion of stoichiometry is the most complex, but if you can do the grams to grams conversion, all of the other types of conversions will follow.
You find out what the molar mass is per mole, for both water and oxygen.
1 mole of any gas has a volume of 22.4 L at 0°C and 1 atm of pressure
Example:
Use this equation 2H2O → 2H2 + O2
You have 20 liters of water. How many grams of oxygen is left over?
To solve this equation, you must use stoichiometry. To convert anything involving liters, instead of using molar mass such as that in a grams related equation, you use 22.4.
Percent Composition- tells the percent, by mass of a particular compound.
Find the molar mass of the compound.
Find the mass of the part of the compound you need.
Determine the percent.
Example:
CO2= 44 g/mol total
C= 12 grams O2= 32 grams
12/44= .27 * 100 = 27%
32/44= .73 * 100 = 73%
Example:
C6H12O6= 180 g/mol total
C6= 72 grams H12= 12 grams O6=96 grams
72/180= .40 * 100 = 40%
12/180 = .07 * 100= 7%
96/180 = .53 * 100 = 53 %
Empirical Formulas- smallest ratio of atoms in a compound.
Find the number of moles of each element in the compound.
Set the elements in a small whole number ratio.
Example:
A compound is 2.8 grams N and 6.4 grams O by mass:
................................................................Reduces to:
2.8 g N 1 mole = .2 mol ...................→..1 mol
......................14 g N
6.4 g O 1 mole = .4 mol ....................→ ..2 mol
......................16 g O
.....................................1:2 ratio N + O2 = NO2
Example:
A compound is 65.4 grams As and 34.6 grams O by mass
.................................................... .......Rounds to:
65.4 g As 1 mole = .87.........................1 mol
........................ 75 g As
34.6 g O 1 mole = 2.7........................ 22.5 mol
.......................16 g O ................................1: 2.5 ratioAs + O = As2O5
Molecular Formula- a chemical formula based on analysis and molecular weight
Find the emperical formula.
Find the mass of the emperical formula.
Divide the molar mass of the molecular formula by the molat mass of the emperical formula.
Multiply the emperical formula by the answer from step 3.
Example:
A compound has a empericial formula of CH and a molar mass of 78 grams.mol. Find the molecular formula.
CH= 13 grams/mol
78/13= 6
CH * 6 = C6H6
Citations:
All content on this page is from notes taken in class or the book.
All grahpics on this page are from the search engine google, then their respectable website.
.......................................................Stoichiometry
Stoichiometry- the calculation of quantitative (measurable) relationships of the reactants and products in a balanced chemical reaction.
Cookie Example:
What if you want 12 dozen cookies?
This example shows that because each total differs, the amount of ingredients needed also differs. This is similar to stoichiometry in the way that to get the total amount needed you need to change the number of each element.
Steps To Stoichiometry:
You can also download our stoichiometry calculator here:
To download, select save, unzip the file, and double click on set up.
This program will not solve your stoichiometry calculations unless you know what you are doing.
It does not use any form of generic A → B equation, instead A = what you have and B = what you want.
Example:
1 mole of any gas has a volume of 22.4 L at 0°C and 1 atm of pressure
Example:
Percent Composition- tells the percent, by mass of a particular compound.
Example:
Example:
Empirical Formulas- smallest ratio of atoms in a compound.
Example:
................................................................Reduces to:
- 2.8 g N 1 mole = .2 mol ...................→. .1 mol
......................14 g N- 6.4 g O 1 mole = .4 mol ....................→ ..2 mol
......................16 g O.....................................1:2 ratio N + O2 = NO2
Example:
.................................................... .......Rounds to:
- 65.4 g As 1 mole = .87.........................1 mol
........................ 75 g As- 34.6 g O 1 mole = 2.7........................ 22.5 mol
.......................16 g O................................ 1: 2.5 ratio As + O = As2O5
Molecular Formula- a chemical formula based on analysis and molecular weight
Example:
Citations:
All content on this page is from notes taken in class or the book.
All grahpics on this page are from the search engine google, then their respectable website.
--- Page made by: Jordan Schaer and Darian Gist